Show that

**perash** · April 21st 2008, 09:15 AM

Show that
cot^2(p/7) + cot^2(2p/7) + cot^2(3p/7) = 5.

**topsquark** · April 21st 2008, 12:41 PM

Originally Posted by perash

Show that
cot^2(p/7) + cot^2(2p/7) + cot^2(3p/7) = 5.

For what value(s) of p? This is not true in general.

-Dan

EDIT: Aaaah. You meant $p = \pi$ .

**JaneBennet** · April 28th 2008, 07:07 PM

See this: http://www.mathisfunforum.com/viewto...d=79371#p79371

It was proved that $\tan^2{\frac{\pi}{7}},\tan^2{\frac{2\pi}{7}},\tan^ 2{\frac{3\pi}{7}}$ are the roots of the equation $x^3-21x^2+35x-7=0$ .

Hence $\cot^2{\frac{\pi}{7}},\cot^2{\frac{2\pi}{7}},\cot^ 2{\frac{3\pi}{7}}$ are the roots of the equation $\left(\frac{1}{x}\right)^3-21\left(\frac{1}{x}\right)^2+35\left(\frac{1}{x}\r ight)-7=0$ , i.e. of the equation $7x^3-35x^2+21x-1=0$ .

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